2025 AMC 10B Problem 4

By place value, $\underline{a}\,\underline{b}$ in base seven is $7a + b,$ and $\underline{b}\,\underline{a}$ in base nine is $9b + a$

The equation reduces to $3a = 4b;$ find digits with $a$ a valid base-seven digit

By place value, $\underline{a}\,\underline{b}$ in base seven is $7a + b,$ and $\underline{b}\,\underline{a}$ in base nine is $9b + a.$ Set them equal: $7a + b = 9b + a,$ so $6a = 8b,$ that is $3a = 4b.$ Thus $a = 4t$ and $b = 3t.$ Because these are two-digit numerals, $t \ne 0;$ because $a$ is a base-seven digit, $t \le 1.$ Hence $a = 4$ and $b = 3,$ so $a + b = 7.$ Therefore, the answer is A.